Use Mean Value Theorem to show |sin(a)-sin(b)|<=|a-b|

[nona.m.nona]

Use the Mean Value Theorem to prove the inequality |sin(a) - sin(b)| <= |a - b| for all a and b.

The MVT says that there is some c between a and b so sin(a) - sin(b) = cos(c)(a - b). Where do I go from there?

[stapel_eliz]

Use the Mean Value Theorem to prove the inequality |sin(a) - sin(b)| <= |a - b| for all a and b.

The MVT says that there is some c between a and b so sin(a) - sin(b) = cos(c)(a - b).

If the two sides are equal, then their absolute values are certainly equal:

. . . . .sin(a) - sin(b) = cos(c)(a - b)

. . . . .|sin(a) - sin(b)| = |cos(c)(a - b)|

You can then split the absolute value on the right-hand side:

. . . . .|sin(a) - sin(b)| = |cos(c)||a - b|

You would like to have the following:

. . . . .|sin(a) - sin(b)| <= |a - b|

Think about the cosine function, and the range of values it takes on. (Hint: |a - b| = (1)|a - b|.)

Eliz.